Question

A pair of concentric spherical conductors of radii a and b are connected by a wire. A point charge q is detached from the inner one and moved radially with uniform velocity v to the outer one. Show that the rate of transfer of the induced charge from the inner to the outer sphere is dQ/dt = -qab(b - a)$^{-1}$v(a + vt)$^{-2}$.

Answer 1

The rate of transfer of the induced charge from the inner to the outer sphere is $dQ/dt = -qab(b - a)^{-1}v(a + vt)^{-2}$.

Answer 2

Let $Q_{inner}$ be the net charge on the inner conductor and $Q_{outer}$ be the net charge on the outer conductor. The potential at the surface of the inner conductor (radius $a$) is given by: $V_a = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_{inner}}{a} + \frac{q}{r} + \frac{Q_{outer}}{b} \right)$ The potential at the surface of the outer conductor (radius $b$) is given by: $V_b = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_{inner}}{b} + \frac{q}{b} + \frac{Q_{outer}}{b} \right)$

Since the conductors are connected by a wire, $V_a = V_b$. Equating the expressions: $\frac{Q_{inner}}{a} + \frac{q}{r} + \frac{Q_{outer}}{b} = \frac{Q_{inner}}{b} + \frac{q}{b} + \frac{Q_{outer}}{b}$ $\frac{Q_{inner}}{a} + \frac{q}{r} = \frac{Q_{inner}}{b} + \frac{q}{b}$ $Q_{inner} \left( \frac{1}{a} - \frac{1}{b} \right) = q \left( \frac{1}{b} - \frac{1}{r} \right)$ $Q_{inner} \left( \frac{b-a}{ab} \right) = q \left( \frac{r-b}{br} \right)$ $Q_{inner} = q \frac{ab}{b-a} \frac{r-b}{br} = q \frac{a(r-b)}{r(b-a)}$

Substituting $r = a+vt$: $Q_{inner}(t) = q \frac{a(a+vt-b)}{(a+vt)(b-a)} = -q \frac{a(b-(a+vt))}{(a+vt)(b-a)}$

The rate of transfer of charge from the inner to the outer sphere is the rate at which the charge on the inner sphere decreases, which is $-\frac{dQ_{inner}}{dt}$. We have $Q_{inner}(t) = -q \frac{a}{b-a} \left( \frac{b}{r(t)} - 1 \right)$. Using the chain rule, $\frac{dQ_{inner}}{dt} = \frac{dQ_{inner}}{dr} \frac{dr}{dt}$. Given $r(t) = a+vt$, so $\frac{dr}{dt} = v$. $\frac{dQ_{inner}}{dr} = -q \frac{a}{b-a} \frac{d}{dr} \left( \frac{b}{r} - 1 \right) = -q \frac{a}{b-a} \left( -\frac{b}{r^2} \right) = \frac{qab}{(b-a)r^2}$. Therefore, $\frac{dQ_{inner}}{dt} = \frac{qab}{(b-a)r^2} \cdot v = \frac{qabv}{(b-a)r^2}$. Substituting $r = a+vt$: $\frac{dQ_{inner}}{dt} = \frac{qabv}{(b-a)(a+vt)^2}$. The rate of transfer of charge from the inner to the outer sphere is $-\frac{dQ_{inner}}{dt}$. $\frac{dQ}{dt} = - \frac{qabv}{(b-a)(a+vt)^2} = -qab(b - a)^{-1}v(a + vt)^{-2}$.